WebMar 31, 2024 · create or replace procedure get_database_ddl (database_name string) returns string language javascript execute as caller as $$ var output = `create or replace database $ {DATABASE_NAME};\n\n`; var query_sch = `select schema_name from snowflake.account_usage.schemata where catalog_name = :1 and deleted is null;`; var … WebAug 21, 2024 · 1 Answer Sorted by: 1 If you are searching for SYNONYM/ALIAS, they are not supported by Snowflake. CREATE SYNONYM EMP FOR USER.EMPLOYEE; Reference: Migrating Oracle Database to Snowflake: Reference Manual - APPENDIX D Depending on requirements VIEW could be used instead: CREATE VIEW USER.EMP …
DE30000 / GDC-1220: Snowflake Receive "ERROR: SQLP
WebOct 6, 2024 · Snowflake does not have something like a ROWID either, so there is no way to identify duplicates for deletion. It is possible to temporarily add a "is_duplicate" column, eg. numbering all the duplicates with the ROW_NUMBER () function, and then delete all records with "is_duplicate" > 1 and finally delete the utility column. WebSnowflake also has built-in support for regular expressions. In SQL Server, you have the (limited) LIKE clause, but true regular expression support in T-SQL is not present. You … mcdonald\\u0027s owen sound
Alias name in Snowflake - Stack Overflow
WebJul 25, 2024 · You cannot join to a subquery outside of the subquery without an alias. Here’s an example of a subquery with an alias and a join outside of itself: select t.*, lookup.name from transactions as t left join (select distinct id, name from duplicate_data) as lookup on t.attribute_id = lookup.id Common Table Expressions (CTEs) WebSep 29, 2024 · flatten is a table function, not a scalar function. It's being called in the SQL as a scalar function rather than a table function, and that's why Snowflake is reporting it can't find the function. If you change this line to use it as a table function, it seems to do what you want: select * from cte, table (flatten (cte.dates)) Share WebMar 31, 2024 · Identify Which IDs in the Table Have Duplicate Records A simple way to list out all of the IDs in the table which have duplicate records is to pair a COUNT (*) aggregation with our ID field using GROUP BY, … mcdonald\u0027s owensville